Divide the following complex numbers. $ \dfrac{-5-25i}{5-i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${5+i}$ $ \dfrac{-5-25i}{5-i} = \dfrac{-5-25i}{5-i} \cdot \dfrac{{5+i}}{{5+i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-5-25i) \cdot (5+i)} {(5-i) \cdot (5+i)} = \dfrac{(-5-25i) \cdot (5+i)} {5^2 - (-1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-5-25i) \cdot (5+i)} {(5)^2 - (-1i)^2} = $ $ \dfrac{(-5-25i) \cdot (5+i)} {25 + 1} = $ $ \dfrac{(-5-25i) \cdot (5+i)} {26} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-5-25i}) \cdot ({5+i})} {26} = $ $ \dfrac{{-5} \cdot {5} + {-25} \cdot {5 i} + {-5} \cdot {1 i} + {-25} \cdot {1 i^2}} {26} $ Evaluate each product of two numbers. $ \dfrac{-25 - 125i - 5i - 25 i^2} {26} $ Finally, simplify the fraction. $ \dfrac{-25 - 125i - 5i + 25} {26} = \dfrac{0 - 130i} {26} = -5i $